Hash算法专题

1、【HDU 3068】最长回文

题意：求一个字符串（len<=110000）的最长回文串

解题思路：一般解法是manacher，但是这一题用hash也是可以ac的

假设当前判断的是以i为中心偶数最长回文串，那么s[2*i+1-k……i]与s[i+1……k]的哈希值必定相同

假设当前判断的是以i为中心奇数最长回文串，那么s[2*i-k……i-1]与s[i+1……k]的哈希值必定相同

用二分求出相应的k

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <string>
 6 using namespace std;
 7 #define ull unsigned long long
 8 const int inf = 110000+10;
 9 const ull bas = 311;
10 ull lhas[inf], rhas[inf], base[inf];
11 char s[inf];
12 int len, ans, even, odd;
13 int getans(int li, int x, int y)
14 {
15     int l = y, r = li;
16     while(l<=r)
17     {
18         int mid = (l+r)/2;
19         int j = x+y-mid;
20         int right = rhas[mid] - rhas[y-1]*base[mid-y+1];
21         int left = lhas[j] - lhas[x+1]*base[mid-y+1];
22         if(left ^ right) r = mid-1;
23         else l = mid+1;
24     }
25     return r-y+1;
26 }
27 int main()
28 {
29     base[0] = 1;
30     for(int i=1; i<inf; i++) base[i] = base[i-1]*bas;
31     while(~scanf("%s", s+1))
32     {
33         s[0] = ‘$‘;
34         int len = strlen(s+1);
35         lhas[len+1] = rhas[0] = 0;
36         for(int i=1; i<=len; i++)
37         {
38             rhas[i] = rhas[i-1]*bas+s[i]-‘a‘;
39         }
40         for(int i=len; i>=1; i--)
41         {
42             lhas[i] = lhas[i+1]*bas+s[i]-‘a‘;
43         }
44         ans = 1;
45         for(int i=1; i<=len; i++)
46         {
47             if(s[i]==s[i+1])
48             {
49                 even = getans(min(2*i, len), i, i+1);
50                 ans = max(ans, 2*even);
51             }
52             if(s[i-1]==s[i+1])
53             {
54                 odd = getans(min(2*i-1, len), i-1, i+1);
55                 ans = max(ans, 2*odd+1);
56             }
57         }
58         printf("%d\n", ans);
59     }
60     return 0;
61 }